Difference between revisions of "2020 AMC 12B Problems/Problem 17"

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How many polynomials of the form <math>x^5 + ax^4 + bx^3 + cx^2 + dx + 2020</math>, where <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> are real numbers, have the property that whenever <math>r</math> is a root, so is <math>\frac{-1+i\sqrt{3}}{2} \cdot r</math>? (Note that <math>i=\sqrt{-1}</math>)
 
How many polynomials of the form <math>x^5 + ax^4 + bx^3 + cx^2 + dx + 2020</math>, where <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> are real numbers, have the property that whenever <math>r</math> is a root, so is <math>\frac{-1+i\sqrt{3}}{2} \cdot r</math>? (Note that <math>i=\sqrt{-1}</math>)
  
==Solution==
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<math>\textbf{(A) } 0 \qquad \textbf{(B) }1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 4</math>
We notice that <math>\frac{1+i\sqrt{3}}{2} = cis (\frac{2\pi}{3})</math>, so in order for every root <math>r</math> to have <math>rcis(\frac{2\pi}{3})</math> as a solution, 3 of the roots of the polynomial must be <math>r</math>, <math>rcis({2\pi}{3})</math>, <math>rcis({4\pi}{3})</math>. However, since the polynomial is degree 5, there must be two additional roots, but in order for each of these roots to reach eachother by multiplying by 120 degrees, there must be three of them, but this clearly cannot be the case. This means that the polynomial is degree 5 with the three previously determined roots and they have a multiplicity. Moreover, by Vieta's, we know that there is only one possible value for r as <math>r^5 = 2020</math>.Therefore, the polynomial is in the form <math>(x-r)^m(x-rcis({2\pi}{3})^n(x-rcis({4\pi}{3})^p</math>. But in order for the coefficients of the polynomial to all be real, <math>n = p</math> due to <math>rcis{2\pi}{3}</math> and <math>rcis{4\pi}{3}</math> being conjugates. Since <math>m+n+p = 5</math> as the polynomial is 5th degree, we have two possible solutions for <math>(m, n, p)</math> which are <math>(1,2,2)</math> and <math>(3,1,1)</math> yielding two possible polynomials. The answer is thus <math>\boxed{C) 2}</math>.
 
  
-- Murtagh
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==Solution 1==
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Let <math>P(x) = x^5+ax^4+bx^3+cx^2+dx+2020</math>. We first notice that <math>\frac{-1+i\sqrt{3}}{2} = e^{i\frac{2\pi}{3}}</math>.  That is because of Euler's Formula : <math>e^{ix} = cos(x) + i \cdot sin(x)</math>. <math>\frac{-1+i\sqrt{3}}{2}</math> = <math>-\frac{1}{2} + i \cdot \frac {sqrt{3}}{2}</math> = <math>cos(120) + i \cdot sin(120)</math> = <math>e^{i \cdot 120(degrees)}</math> = <math>e^{i \frac{2}{3} \pi (radians)}</math>.  In order <math>r</math> to be a root of <math>P</math>, <math>re^{i\frac{2\pi}{3}}</math> must also be a root of P, meaning that 3 of the roots of <math>P</math> must be <math>r</math>, <math>re^{i\frac{2\pi}{3}}</math>, <math>re^{i\frac{4\pi}{3}}</math>. However, since <math>P</math> is degree 5, there must be two additional roots. Let one of these roots be <math>w</math>, if <math>w</math> is a root, then <math>we^{i\frac{2\pi}{3}}</math> and <math>we^{i\frac{4\pi}{3}}</math> must also be roots. However, <math>P</math> is a fifth degree polynomial, and can therefore only have <math>5</math> roots. This implies that <math>w</math> is either <math>r</math>, <math>re^{i\frac{2\pi}{3}}</math>, or <math>re^{i\frac{4\pi}{3}}</math>. Thus we know that the polynomial <math>P</math> can be written in the form <math>(x-r)^m(x-re^{i\frac{2\pi}{3}})^n(x-re^{i\frac{4\pi}{3}})^p</math>. Moreover, by Vieta's, we know that there is only one possible value for the magnitude of <math>r</math> as <math>||r||^5 = 2020</math>, meaning that the amount of possible polynomials <math>P</math> is equivalent to the possible sets <math>(m,n,p)</math>. In order for the coefficients of the polynomial to all be real, <math>n = p</math> due to <math>re^{i\frac{2\pi}{3}}</math> and <math>re^{i\frac{4\pi}{3}}</math> being conjugates and since <math>m+n+p = 5</math>, (as the polynomial is 5th degree) we have two possible solutions for <math>(m, n, p)</math> which are <math>(1,2,2)</math> and <math>(3,1,1)</math> yielding two possible polynomials. The answer is thus <math>\boxed{\textbf{(C) } 2}</math>.
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~Murtagh
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==Solution 2==
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Let <math>x_1=r</math>, then <math>x_2=(-1+i\sqrt{3})/2 r</math>, x_3=〖((-1+i√3)/2)〗^2 r=(-1-i√3)/2 r, x_4=〖((-1+i√3)/2)〗^3 r=r which means x_4 will be back to x_1
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Now we have 3 different roots of the polynomial, x_(1  ) 〖,x〗_2, and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of the different root is wrong.
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The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2
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~Yelong_Li
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==Video Solution==
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Link: https://www.youtube.com/watch?v=8V5l5jeQjNg
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==See Also==
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{{AMC12 box|year=2020|ab=B|num-b=16|num-a=18}}
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{{MAA Notice}}

Revision as of 19:46, 4 September 2020

Problem

How many polynomials of the form $x^5 + ax^4 + bx^3 + cx^2 + dx + 2020$, where $a$, $b$, $c$, and $d$ are real numbers, have the property that whenever $r$ is a root, so is $\frac{-1+i\sqrt{3}}{2} \cdot r$? (Note that $i=\sqrt{-1}$)

$\textbf{(A) } 0 \qquad \textbf{(B) }1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 4$

Solution 1

Let $P(x) = x^5+ax^4+bx^3+cx^2+dx+2020$. We first notice that $\frac{-1+i\sqrt{3}}{2} = e^{i\frac{2\pi}{3}}$. That is because of Euler's Formula : $e^{ix} = cos(x) + i \cdot sin(x)$. $\frac{-1+i\sqrt{3}}{2}$ = $-\frac{1}{2} + i \cdot \frac {sqrt{3}}{2}$ = $cos(120) + i \cdot sin(120)$ = $e^{i \cdot 120(degrees)}$ = $e^{i \frac{2}{3} \pi (radians)}$. In order $r$ to be a root of $P$, $re^{i\frac{2\pi}{3}}$ must also be a root of P, meaning that 3 of the roots of $P$ must be $r$, $re^{i\frac{2\pi}{3}}$, $re^{i\frac{4\pi}{3}}$. However, since $P$ is degree 5, there must be two additional roots. Let one of these roots be $w$, if $w$ is a root, then $we^{i\frac{2\pi}{3}}$ and $we^{i\frac{4\pi}{3}}$ must also be roots. However, $P$ is a fifth degree polynomial, and can therefore only have $5$ roots. This implies that $w$ is either $r$, $re^{i\frac{2\pi}{3}}$, or $re^{i\frac{4\pi}{3}}$. Thus we know that the polynomial $P$ can be written in the form $(x-r)^m(x-re^{i\frac{2\pi}{3}})^n(x-re^{i\frac{4\pi}{3}})^p$. Moreover, by Vieta's, we know that there is only one possible value for the magnitude of $r$ as $||r||^5 = 2020$, meaning that the amount of possible polynomials $P$ is equivalent to the possible sets $(m,n,p)$. In order for the coefficients of the polynomial to all be real, $n = p$ due to $re^{i\frac{2\pi}{3}}$ and $re^{i\frac{4\pi}{3}}$ being conjugates and since $m+n+p = 5$, (as the polynomial is 5th degree) we have two possible solutions for $(m, n, p)$ which are $(1,2,2)$ and $(3,1,1)$ yielding two possible polynomials. The answer is thus $\boxed{\textbf{(C) } 2}$.

~Murtagh

Solution 2

Let $x_1=r$, then $x_2=(-1+i\sqrt{3})/2 r$, x_3=〖((-1+i√3)/2)〗^2 r=(-1-i√3)/2 r, x_4=〖((-1+i√3)/2)〗^3 r=r which means x_4 will be back to x_1 Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2, and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of the different root is wrong. The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2

~Yelong_Li

Video Solution

Link: https://www.youtube.com/watch?v=8V5l5jeQjNg

See Also

2020 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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All AMC 12 Problems and Solutions

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